Gravitation Solved Examples (11 Physics)
Gravitation is the force of attraction between masses. These examples cover Newton's law of universal gravitation, gravitational field, orbital motion, and
TL;DR: Gravitation is the force of attraction between masses. These examples cover Newton's law of universal gravitation, gravitational field, orbital motion…
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Gravitation is the force of attraction between masses. These examples cover Newton's law of universal gravitation, gravitational field, orbital motion, and
Gravitation — Solved Numerical Examples (Step by Step)
Example 1: Calculate the gravitational force between two masses of 10 kg and 5 kg separated by 2 m. (G = 6.67 × 10^-11 N⋅m²/kg²)
Solution: Use Newton's law of universal gravitation:
F = G × (m1 × m2) / r²
Given: m1 = 10 kg, m2 = 5 kg, r = 2 m, G = 6.67 × 10^-11 N⋅m²/kg²
F = (6.67 × 10^-11) × (10 × 5) / (2)²
F = (6.67 × 10^-11) × 50 / 4
F = (6.67 × 10^-11) × 12.5
F = 83.375 × 10^-11 N
F = 8.34 × 10^-10 N
Example 2: The gravitational field strength at Earth's surface is 10 N/kg. Find the gravitational force on a 50 kg person standing on Earth.
Solution: Gravitational force = mass × gravitational field strength
F = m × g
Given: m = 50 kg, g = 10 N/kg
F = 50 × 10 = 500 N
This is the weight of the person.
Example 3: If the distance between two objects is doubled, how does the gravitational force change?
Solution: From F = G × (m1 × m2) / r²
If r becomes 2r:
F_new = G × (m1 × m2) / (2r)²
F_new = G × (m1 × m2) / (4r²)
F_new = (1/4) × [G × (m1 × m2) / r²]
F_new = F_original / 4
The force becomes 1/4 of the original (decreases by a factor of 4).
Example 4: Calculate the acceleration due to gravity on the surface of a planet with mass M = 6 × 10^24 kg and radius R = 6.4 × 10^6 m. (G = 6.67 × 10^-11 N⋅m²/kg²)
Solution: Acceleration due to gravity:
g = G × M / R²
Given: G = 6.67 × 10^-11, M = 6 × 10^24 kg, R = 6.4 × 10^6 m
g = (6.67 × 10^-11) × (6 × 10^24) / (6.4 × 10^6)²
g = (40.02 × 10^13) / (40.96 × 10^12)
g = (40.02 × 10^13) / (40.96 × 10^12)
g = 9.77 m/s² ≈ 9.8 m/s²
Example 5: A satellite orbits Earth at a height where g = 2.5 m/s². What is the gravitational force on a 1000 kg satellite?
Solution: At the satellite's altitude, gravitational field strength g = 2.5 m/s²
Gravitational force = m × g
F = 1000 × 2.5 = 2500 N
This force provides the centripetal force needed for circular orbit.
Example 6: Earth's mass is 6 × 10^24 kg and its radius is 6.4 × 10^6 m. Calculate the escape velocity from Earth. (G = 6.67 × 10^-11 N⋅m²/kg²)
Solution: Escape velocity is given by:
v_e = √(2GM / R)
Given: G = 6.67 × 10^-11, M = 6 × 10^24 kg, R = 6.4 × 10^6 m
v_e = √(2 × 6.67 × 10^-11 × 6 × 10^24 / 6.4 × 10^6)
v_e = √(80.04 × 10^13 / 6.4 × 10^6)
v_e = √(12.5 × 10^7)
v_e = √(1.25 × 10^8)
v_e ≈ 11.2 × 10^3 m/s = 11.2 km/s
Example 7: The Moon orbits Earth at an average distance of 3.84 × 10^8 m. If Earth's mass is 6 × 10^24 kg, find the period of Moon's orbit. (G = 6.67 × 10^-11 N⋅m²/kg²)
Solution: From Kepler's third law: T² = (4π² / GM) × r³
T = 2π × √(r³ / GM)
Given: r = 3.84 × 10^8 m, G = 6.67 × 10^-11, M = 6 × 10^24 kg
T = 2π × √((3.84 × 10^8)³ / (6.67 × 10^-11 × 6 × 10^24))
T = 2π × √(5.65 × 10^25 / 4 × 10^14)
T = 2π × √(1.41 × 10^11)
T = 2π × 3.75 × 10^5 seconds
T ≈ 2.36 × 10^6 seconds ≈ 27.3 days
Tips
- Gravitational force is always attractive and acts along the line joining the centers of mass.
- Gravitational field strength g represents the force per unit mass (units: N/kg or m/s²).
- Doubling distance reduces force to 1/4; tripling distance reduces force to 1/9 (inverse square law).
- Escape velocity is independent of the object's mass; it depends only on the planet's mass and radius.
Frequently Asked Questions
Why is the gravitational force so weak compared to other forces?
The gravitational constant G is extremely small (6.67 × 10^-11), making gravitational force weak at everyday scales. However, it becomes significant for large masses like planets and stars where it dominates over other forces.
How does gravitational potential energy differ from kinetic energy in orbit?
In a stable orbit, gravitational potential energy and kinetic energy are related such that total mechanical energy is constant. As an object falls closer to a massive body, potential energy decreases while kinetic energy increases.
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